
The ground state of the hydrogen atom is \[13.6eV\] . When its electron is in the first excited state? Its excitation energy is:
A.\[10.2eV\]
B.Zero
C.\[3.4eV\]
D.\[6.8eV\]
Answer
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Hint:
In this question first calculate the energy of an electron present in the first excited state of the Hydrogen atom with Bohr’s expression to calculate energy in nth orbit. Now after calculating the energy in the first excited state that is the second orbit, take the difference of energy when the electron is in the ground state and energy of the first excited state that we’ve just calculated.
Complete step by step answer:
According to Bohr's model, an electron would absorb energy within the sort of photons to urge excited to a better energy state as long as the photon's energy was equal to the energy difference between the initial and final energy levels. After jumping to the upper energy level—also called the excited state—the excited electron would be during a less stable position, so it might quickly emit a photon to relax back to a lower, more stable energy level.
The energy of an electron in Bohr’s orbit of hydrogen atom is given by the expression:
\[{E_n} = \dfrac{{ - 13.6{z^2}}}{{{n^2}}}\] ------(1)
Where:
\[z\] is the atomic number.
\[n\] is the excited state of an electron.
For Hydrogen: \[z = 1\]
And we are the energy of Hydrogen atom in ground state:
\[{E_1} = - 13.6eV\]
For the energy of atom in first excited state:
\[n{\text{ }} = {\text{ }}2,\]
Putting the values in equation and calculating we get the value of energy as:
\[{E_2} = - 3.4eV\]
Now taking the difference of the two energies to know how much change energy is needed to go from ground state to first excited state:
\[\Delta E = {E_2} - {E_1}\]
Putting above calculated values:
\[\Delta E = - 3.4 - ( - 13.6) = 10.2eV\]
Hence, Option A is correct.
Note:
As we know energy is defined as zero when the electron is an infinite distance from the nucleus. When the electron comes closer, the energy of the electron decreases, so the energy of the system becomes negative. Even for a Hydrogen atom in an excited state, the energy is negative, but it is less negative than for an atom in the ground state.
In this question first calculate the energy of an electron present in the first excited state of the Hydrogen atom with Bohr’s expression to calculate energy in nth orbit. Now after calculating the energy in the first excited state that is the second orbit, take the difference of energy when the electron is in the ground state and energy of the first excited state that we’ve just calculated.
Complete step by step answer:
According to Bohr's model, an electron would absorb energy within the sort of photons to urge excited to a better energy state as long as the photon's energy was equal to the energy difference between the initial and final energy levels. After jumping to the upper energy level—also called the excited state—the excited electron would be during a less stable position, so it might quickly emit a photon to relax back to a lower, more stable energy level.
The energy of an electron in Bohr’s orbit of hydrogen atom is given by the expression:
\[{E_n} = \dfrac{{ - 13.6{z^2}}}{{{n^2}}}\] ------(1)
Where:
\[z\] is the atomic number.
\[n\] is the excited state of an electron.
For Hydrogen: \[z = 1\]
And we are the energy of Hydrogen atom in ground state:
\[{E_1} = - 13.6eV\]
For the energy of atom in first excited state:
\[n{\text{ }} = {\text{ }}2,\]
Putting the values in equation and calculating we get the value of energy as:
\[{E_2} = - 3.4eV\]
Now taking the difference of the two energies to know how much change energy is needed to go from ground state to first excited state:
\[\Delta E = {E_2} - {E_1}\]
Putting above calculated values:
\[\Delta E = - 3.4 - ( - 13.6) = 10.2eV\]
Hence, Option A is correct.
Note:
As we know energy is defined as zero when the electron is an infinite distance from the nucleus. When the electron comes closer, the energy of the electron decreases, so the energy of the system becomes negative. Even for a Hydrogen atom in an excited state, the energy is negative, but it is less negative than for an atom in the ground state.
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